5. PHASES AND SOLUTIONS n Thermodynamics of Vapor Pressure 5.. At equilibrium, G(graphite) G(diamond); i.e., G 2 0. We are given G 2900 J mol. ( G/ P) T V V 2.0 g mol.95 0 6 m 3 mol Holding T constant 3.5 2.25 0 6 m 3 2 d G P 2 P VdP G 2 G V(P 2 P ) P 2 G 2 G V + P 5.2. V l 0 2900 J mol.92 0 6 m 3 mol + 05 Pa.5 0 9 Pa or.5 0 4 bar At this pressure the system must also be at approximately 000 K. Molar mass density 8.0 0 3 kg mol 0.958 kg dm 3 8.80 0 3 dm 3 mol V v 8.0 0 3 kg mol 5.98 0 4 kg dm 3 30.2 dm3 mol From Eq. 5.8, dp dt P T S vap V v V l 08.72 J K (30.2 0.088) dm 3 mol 3.62 J dm 3 K P 3.62 J dm 3 K ( K) 3.62 0 3 J m 3
96 n CHAPTER 5 Since J kg m 2 s 2 or N m and Pa kg m s 2 or N m 2 P 3.62 0 3 Pa 5.3. ln P 2 P vap H m T 2 T R vap H m T 2 T T 2 T R ln 0.325 3.7 vap H m /J mol 373.5 298.5 373.5 298.5 vap H m /J mol 3.4646 (8.345) 254 75 42.73 kj mol vap H m 42.7 kj mol 8.345 The CRC Handbook gives 40.57 kj mol. 5.4. Using the Clausius-Clapeyron equation, we get P 40 656 ln 0.00603 8.345 273.6 373.5 Therefore, P 2 0.00603 atm e 4.7967 0.73029 atm. 5.5. The molar volume of iodine, I 2, is V m Then from Eq. 5.23, 2 4.7967. 0.253 8 kg mol 4.93 kg dm 3 5.48 0 3 dm 3 mol 5.48 0 6 m 3 mol ln Pg P g 2 V m (P P 2 ) RT ln Pg P g 2 5.48 0 6 (0.3 0 6 0 300) 8.345 33.5 2.00 P g /Pg 2 7.40 At 0.3 kpa the pressure is 33 Pa. Therefore, at 0.3 0 3 kpa the pressure is 7.40 33 Pa 984 Pa.
PHASES AND SOLUTIONS n 97 5.6. The cubic expansion coefficient is a second-order transition since it can be expressed as α V T G P T P α T 5.7. From Eq. 5.6, vap H m T P 2 T R ln 2 P vap H m /J mol (4.975 0 4 ) 8.345 ln 3.86.94 (200) (8.345) (2.7987) 46 822 vap H m 46 773 J mol 46.8 kj mol 5.8. ln P 2 P vap H m T 2 T R vap H m (T 2 T ) R T 2 T P 2 4 Torr/760 Torr atm 0 325 Pa atm 866.5 Pa At T 286 + 273.5 559.5 K P 0 325 Pa At T 2 45 + 273.5 48.5 P 2 866.5 Pa ln 866.5 0 325 vap H m /J mol 8.345 48.5 559.5 (48.5) (559.5) vap H m /J mol (8.345) (233 809) +3.994 4 55.07 kj mol The CRC Handbook value is 7.02 kj mol. The error is large, but considering the relative molecular mass of the compound, its high boiling point, and the wide range of T and P involved in the calculation, it is not surprising that the error is so large.
98 n CHAPTER 5 5.9. Using Trouton s rule, vap H m 88 J K mol 342.0 K 30.05 kj mol % error 3.92 30.05 3.92 00 5.7% 5.0. Applying Trouton s rule, vap H T vap S (88 J K mol )(383.77 K) ln P 2 P 33.8 kj mol vap H T T 2 R ln P 2 atm 33.8 kj mol P 2 383.77 8.345 J K mol 353.5 0.985 0.399 atm 5.. 2-Propanone is not particularly associated. Therefore, using T 9.3 0 4 T b P, we have T 9.3 0 4 (329.35) (2.825) 0.8653 K Thus T b at 98.5 kpa is 329.35 0.87 328.48 K 5.2. At the triple point, the two vapor pressures must be equal since the liquid, solid, and vapor are all in equilibrium with each other. Therefore, we set 266 T trp + 22.76 3775 T trp + 26.88. Solving for T trp, we obtain the triple point temperature: T trp 265.5 K. Substituting the triple point temperature into one of the two equations given, we calculate the triple point pressure: ln P l 266 265.5 + 22.76 2.737. P l 3.42 0 5 Pa. 5.3. Since water is associated, the numerical value 7.5 0 4 should be used in Eq. 5.20. T b 373.5 K and P 02.7 0.3.4 kpa, and thus T 7.5 0 4 373.5.4 0.39 K The calculated boiling point at 0.3 kpa is thus 373.52 0.39 373.3 K, an error of only 0.02 K. 5.4. Molar volume of water 8.05 g mol /0.996374 g cm 3 8.08 cm 3 mol
PHASES AND SOLUTIONS n 99 ln P/Torr 27.536 8.08 cm 3 mol (760.000 Torr 27.536 Torr) 82.056 cm atm K (760 Torr atm ) (273.5 + 27.5) 8.08(732.46) 8 749 000 7.063 0 4 ln P/Torr 7.063 0 4 + ln 27.536 7.063 0 4 + 3.355 3.36 P 27.55 Torr This is a rather small correction but may be necessary for very accurate work. 5.5. From Eq. 3.5, dg VdP SdT. Recognize that a change of state occurs at constant temperature. Therefore, differentiating Eq. 3.5 with respect to P at constant temperature, we obtain (see Eq. 3.9) G P T V. Now, for a change of state where G f G i G, there will be a corresponding change in volume, G P T V. 5.6. We start with the Clapeyron equation, Eq. 5.9, and substitute RT + M P rearrangement, Since dp P vap H m dt T(RT + M) T(RT + M) MT R M(RT + M) dp P vap H m dt M T R M Integration with vap H m constant gives vap H m dt (RT + M) for V; this gives, after ln P 2 P vap H m M ln T 2 ln P 2 P vap H m M ln T 2 T vap H m M T RT + M RT 2 + M 5.7. Molar mass of C 2 H 5 OH 46.0695 g mol ln RT 2 + M RT + M Molar volume of C 2 H 5 OH 46.0695 g mol 0.7767 g dm 3 59.34 cm mol 59.34 0 6 m 3 mol
00 n CHAPTER 5 Using Eq. (5.23), we get P/Pa ln 0 325 Pa 00 Torr 760 Torr 59.34 0 6 m 3 mol 00 Torr (0 325 Pa) 8.345 J mol K 308.5 K.000 0 7 Pa 760 Torr ln P/Pa 0.239 + ln.3332 0 4 0.239 + 9.49792 9.729.67996 0 4 Pa.67996 04 Pa 0 325 Pa 760 Torr 26 Torr 5.8. The volume change is V 3.555 2.2670 cm3 g 0.567 0 3 m 3 kg. From the expression derived in Problem 5.5, we have d G VdP. Integrating both sides, we obtain 2 P2 P d G V dp. Let state be the standard state and state 2 be at equilibrium. Then, the expression is G G V(P P ). Now, at equilibrium, the Gibbs energy change G is zero. Also, using 0 5 Pa for the standard pressure P, we get 240 kj kg 0.567 0 3 m 3 kg (P 0 5 ) Pa. Solving for P gives P.53 0 9 Pa. 5.9. From Raoult s law (Eq. 5.26) we find, for toluene, P tol x tol P tol * 0.60 0.85 bar 0. bar. and for benzene, P ben x ben P ben * 0.40 0.53 bar 0.205 bar. Therefore, the total pressure is P tot 0.bar + 0.205 bar 0.36 bar. 5.20. The molar heat of vaporization is vap H m 80 J g 62.058 g mol 49 708 J mol P2 vaph The Clausius-Clapeyron equation is ln. Substituting, we get P R T2 T
PHASES AND SOLUTIONS n 0 760 49 708 ln, 30 8.345 470 T which yields, upon simplification, ln 25.3 5978 470 T or 3.23 5978(2.28 0 3 /T ), from which, we get T 375 K or 02 C. n Raoult s Law, Equivalence of Units, and Partial Molar Quantities 5.2. P T 0.60(3.572) + 0.40(9.657) 2.43 + 3.863 6.006 kpa xvap toluene 2.43 6.006 0.357 5.22. The molality m 2 is the amount of solute divided by the mass of solvent. If W is the mass of solvent, the solution contains m 2 W mol of solute and W /M mol of solvent. The mole fraction is thus m 2 W m 2 M x 2 W /M + m 2 W + m 2 M If we divide each term by its SI unit, (m 2 /mol kg ) (M /kg mol ) x 2 + (m 2 /mol kg ) (M /kg mol ) The customary unit for molar mass M is g mol, and we then obtain (m 2 /mol kg )(M /000 g mol ) x 2 + (m 2 /mol kg )(M /000 g mol ) (m 2 /mol kg )(M /g mol ) 000 + (m 2 /mol kg )(M /g mol ) If the solution is sufficiently dilute, the expression approximates to x 2 (m 2 /mol kg ) (M /g mol )/000 5.23. Rearranging Eq. 5.22, and substituting gives
02 n CHAPTER 5 W 2 M 2 fus TW K f 0.06202 kg mol (20.0 K) (.00 kg).86 K kg mol 0.667 kg. Converting the mass to a volume using density, and assuming that.00 kg of water has a volume of.00 dm 3, we get V 2 0.667 kg/(.088 kg dm 3 ) 0.602 dm 3. So, the volume ratio is 0.602 dm 3 antifreeze to.0 dm 3 water (or approximately 3:5 volume ratio). The elevation of boiling point for this solution will be vap T K b m 2 0.5 (0.667/0.06202) 5.48 K. This means that the solution will boil at 378.6 K or 05 C. 5.24. If V is the volume of solution, concentration of solute c 2 ; amount of solute Vc 2 ; mass of solute Vc 2 M 2 where M 2 is the molar mass of solute. Mass of solution Vρ; mass of solvent Vρ Vc 2 M 2 ; amount of solvent (Vρ Vc 2 M 2 )/M. The mole fraction is therefore x 2 Vc 2 (Vρ Vc 2 M 2 )/M + Vc 2 c 2 M ρ + c 2 (M M 2 ) Dividing each term by its SI unit, (c 2 /mol m 3 ) (M /kg mol ) x 2 ρ/kg m 3 + (c 2 /mol m 3 ) [(M M 2 )/kg mol ] The more customary units are, for concentration, mol dm 3 ; for molar mass, g mol ; for density, kg dm 3 g cm 3. Then (000 c 2 /mol dm 3 ) (M /000 g mol ) x 2 (M 000 ρ/g cm 3 + (000 c 2 /mol dm 3 M 2 ) ) 000 g mol (c 2 /mol dm 3 ) (M /g mol ) 000 ρ/g cm 3 + (c 2 /mol dm 3 ) (M M 2 ) g mol If the solution is sufficiently dilute, the density of the solution is approximately that of the pure solvent, ρ, and the second term in the denominator can be neglected. x 2 c 2 M /ρ (c 2 /mol dm 3 ) (M /g mol ) 000 ρ /g cm 3 5.25. It is difficult to work with the two laws in their normal form. Cast Henry s law into the form µ 2 µ 0 2 + RT ln x 2 using the arguments of Section 5.7 and Eq. 5.0. Then
PHASES AND SOLUTIONS n 03 dµ 2 /dx 2 RT/x 2 dµ 2 RT x 2 dx 2 Using the Gibbs-Duhem equation, x dµ + x 2 dµ 2 0 dµ x 2 dµ 2 /x RT/x dx 2 Since x + x 2, dx 2 dx dµ RT/x dx Integration gives µ RT ln x + constant Since x as µ goes to µ 0 µ RT ln x + µ 0 5.26. Using Eq. 5.2, we get fus T.86 K kg mol.0 mol kg.86 K, which is not what is observed. NaCl completely ionizes in solution. Since colligative properties are, to a large extent, determined by the number of particles in solution rather than the actual identity of the species, we should realize that.0 mol of solid NaCl results in a solution containing 2.0 mol of particles. Therefore, we calculate that the temperature change is fus T.86 K kg mol 2.0 mol kg 3.72 K, which accounts for the observation. 5.27 From Problem 5.24, amount of solute Vc 2 ; mass of solvent Vρ Vc 2 M 2 Thus, the molality is m 2 c 2 ρ c 2 M 2 and the concentration in terms of m 2 is m 2 ρ c 2 + m 2 M 2 Dividing throughout by SI units, (m c 2 /mol m 3 2 /mol kg ) (ρ/kg m 3 ) + (m 2 /mol kg ) (M 2 /kg mol )
04 n CHAPTER 5 In terms of more usual units, 000 c 2 /mol dm 3 (m 2 /mol kg ) (000 ρ/g cm 3 ) g M 2 /000 mol + (m 2 /mol kg ) 000(m c 2 /mol dm 3 2 /mol kg ) (ρ/g cm 3 ) 000 + (m 2 /mol kg ) (M 2 /g mol ) In dilute solution, c 2 /mol dm 3 (m 2 /mol kg ) (ρ /g cm 3 ) where ρ is the density of the solvent. For aqueous solutions ρ g cm 3 and therefore the numerical values of the concentration and the molality, in the above units, are very similar. 5.28. Assuming that this is an ideal solution, we use Raoult s law with P Hg ( x M )P * Hg, where P * Hg 768.8 Torr at the boiling point, and P Hg 754. Torr. Therefore, we calculate x M P Hg P * Hg.92 0 2. Now, since.92 0 2.52/MW (.52/MW + 00.0/200.59), we can solve for the molecular mass of the element, which gives MW 8.54 g mol. Comparison with the masses of elements shows that the metal is tin. 5.29. From Eq. 5.3, V NaCl /cm 3 mol V n V NaCl n H2 m O From Eq. 5.37, 7.823 +.74782 m 0.4675 m 2 n NaCl m 55.508 dv NaCl dv H2 O n dv NaCl H2 O Integration from m 0 to m gives dv H 2 O.74782 55.508 m 0.28335 (55.508) m2 dm V H2 O V * H 2 O.74782 2(55.508) m2 + 0.28335 3(55.508) m3 With V * H 2 O 8.068 cm3, V H2 O /cm3 mol 8.068 0.05 744 m 2 + 0.00 706 m 3
PHASES AND SOLUTIONS n 05 5.30. We first develop an expression for ( ρ/ n 2 )n. Since x 2 n 2 /(n + n 2 ), x 2 n 2 n and n (n + n 2 ) 2 ρ dρ n 2 n dx 2 x 2 n 2 n Substitution and division by (n + n 2 ) gives V 2 M 2 ρ (M x + M 2 x 2 ) x ρ 2 dρ dx 2 n (n + n 2 ) 2 dρ dx 2 5.3. Substitution of x 2 0.00 into the expression in Problem 5.23 gives ρ 0.970 609 kg dm 3. Differentiating the ρ equation with respect to x 2 gives dρ dx 2 0.289 30 + 0.598 4 x 2.826 28 x 2 2 + 2.377 52 x3 2.029 05 x4 2 0.245 47 kg dm 3 with M (H 2 O) 0.08 06 kg mol and M 2 (methanol) 0.032 043 kg mol, substitution with x 0.900 gives V 2 0.037 56 dm 3 mol. 5.32. Using Dalton s Law of partial pressures, P c y c P tot, where y c is the mole fraction of chloroform in the vapor phase. P c ( 0.88)220.5 40. Torr. Activity and activity coefficient: a c P c / P * c 40.3/22.8 0.8; f c a c /x c 0.8/( 0.73) 0.630. 5.33. Using Eq. 5.80, 56.8 Torr 55.24 Torr 56.8 Torr 2.5 g (46.0695 g mol ) M 2 (520.8 g) 0.0673.0577 g mol /M 2 M 2 66. g mol
06 n CHAPTER 5 5.34. We prepare a graph of the data from which we see that the solution exhibits positive deviation from Raoult s law. P tot 0 00 90 P iso P benz Pressure/Torr 80 70 60 50 40 30 20 0 0 0.0 0. 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9.0 Mole fraction of Isopropanol From this graph, we construct the following table. a i P i /P * i, f i a i /x i. x I P I (Torr) a I P I 44.0 f I P I x I P B (Torr) a B P B 94.4 f B P B x B 0.20 25.0 0.568 2.84 84.5 0.895.2 0.50 30.0 0.682.36 76.0 0.805.6 0.80 39.0 0.886. 50.0 0.530 2.65 5.35. From Raoult s law, we have for propylene dibromide (p), P p x(l)p * p 0.600(28) Torr (33.33)Pa/Torr 0.239 kpa For ethylene dibromide (e), P e x(l)p * e 0.40(72 Torr) (33.32)Pa/Torr 9.72 kpa The total pressure is P total 0.239 kpa + 9.72 kpa 9.4 kpa. The mole fraction in the vapor phase is given by x P P total
PHASES AND SOLUTIONS n 07 For propylene dibromide, 0.239 kpa x p (v) 9.4 kpa 0.527 For ethylene dibromide, x e (v) 9.72 kpa 9.4 kpa 0.473 5.36. First, calculate the moles of each present. For A, n 2 75.0 g/89.5 g/mol 0.83799 mol. For B, n 000 g/85 g mol 5.4054 mol. Then mole fractions are calculated. For A, x 2 0.83799/6.24339 0.342. For B, x 5.4054/(0.83799 + 5.4054) 0.8658. P total 520 Torr 430 Torr + P 2 P 2 90 Torr Then, from Henry s law, P 2 k x 2 k 90 Torr 0.342 67 Torr 5.37. a. Henry s law applies to the individual gas components, and we require the partial pressures of N 2 and O 2. Since the partial pressure is directly proportional to the mole fraction, for N 2 we have P N2 0.80 atm, and for O 2 we have P O2 0.20 atm. From Eq. 5.28, x(n 2 ) x(o 2 ) P N2 0.80 atm k N2 7.58 0 4 atm P O2 0.20 atm k O2 3.88 0 4 atm.06 0 5 5.5 0 6 b. Since the mole fractions are so small, if we use mol of water as a reference, then there are (to a very good approximation).06 0 5 mol of N 2 and 5.5 0 6 mol of O 2. The concentration calculation requires the volume of the solution, which is obtained from the density: c(n 2 ) c(o 2 ).06 0 5 mol (N 2 ) mol(h 2 O)[0.08 kg (H 2 O)/mol(H 2 O)] 0.9982 kg H 2 O dm 3 5.88 0 4 M 5.5 0 6 mol (O 2 ) mol(h 2 O)[0.08 kg (H 2 O)/mol(H 2 O)] 0.9982 kg H 2 O dm 3 2.86 0 4 M
08 n CHAPTER 5 5.38. Assume that the total vapor pressure of pure benzene is present in the total pressure of 750.0 Torr. P P T P 750.0 94.6 655.4 Torr P 2 k x 2 and k 4.27 0 5 Torr 655.4 Torr 4.27 0 5 Torr x 2 x 2.535 0 3 x 0.99846 Using Raoult s law P x P * 0.99846 94.6 94.45 Thus, the assumption is valid. Next, determine m 2. In 000 g of benzene, n 000 78. 2.80 x 2 n 2 n 2 + 2.80.535 0 3 Solving for n 2 gives 0.0200 mol. Therefore, the molality is 0.0200 m. n Thermodynamics of Solutions 5.39. From Raoult s law, x P /P * and x 2 x (P /P* ), from which we get x 2 P* P P * 0.040 2.332 W 2 /M 2 (W /M ) + (W 2 /M 2 ) 8.04/M 2 0.076 5.55 + 8.04/M 2 8.04/M 2 00.0/8.05 + 8.04/M 2 8.04 0.32 M 2 0.0977 8.4 g mol. The correct value is 82.8 g mol.
PHASES AND SOLUTIONS n 09 5.40. The vapor pressure has been reduced from 40.00 to 26.66 kpa, so that the mole fraction of the solvent is given by Raoult s law as x P 26.66 kpa P * 40.00 kpa 2 3 Let the amount of solute be n 2 mol; since n mol, the mole fraction of solvent is x + n 2 2 3 Consequently, n 2 /2. Since 0.80 kg is half a mole, the molar mass is 0.60 kg mol 60 g mol. 5.4. From Eq. 5.78, P * P x 2 P* 3.3 2.6 x 2 (3.3) x 2 0.053 From Eq. 5.79, x 2 W 2 /M 2 W /M + W 2 /M 2 (M 2 W /M W 2 ) + M 2 W M W 2 x 2 8.0; M 2 M 8.0.00 0.00.80 5.42. f T K f m 2.5 K.5 7.0 m 2 /mol kg ; m 2 0.243 mol kg 0.243 00 mol % impurity 0.243 + 000/28 2.43 8.027 2.7% 00.0% 2.7% 97.3 mol % pure 5.43. a 6.677 9.657 0.694 a 2.24 3.572 0.3399 γ 0.694 0.670.03 γ 2 0.3399 0.330.03 5.44. Amount of NaCl.5 g/58.5 g mol 0.97 mol Amount of H 2 O 00.0 g/8.05 g mol 5.55 mol x n n + n 2 5.55 5.748 0.966, x 2 0.034 a 95.325 0.325 0.94 γ 0.94 0.965 0.975
0 n CHAPTER 5 5.45. From Figure 5.3, the maximum value of the entropy is 5.76 J K mol. Then mix G id 300(5.76) 728 J mol. Therefore, the Gibbs energy of mixing ranges from 0 to.73 kj mol for a 50/50 mixture. Since this is a rather small driving force, in a nonideal solution where mix H 0, the value of mix H must be negative or only slightly positive for mixing to occur. 5.46. a H2 O P H2 O P * H 2 O 2.269 kpa 2.339 kpa 0.970. Since a H2 O γ H 2 O x H 2 O γ H 2 O a H2 O x H2 O 0.970 0.990 0.980 5.47. Find the value of µ A such that µ A + µ B ( G) is equal to the expression given in the problem. ln x A + n A n B a. µ A G n A n B,T,P µ* A + RT ln x A + RTnA RTn B Since and ln x B + Cn B (n A + n B ) n A n B (n A + n B ) 2 Cn A n B (n A + n B ) 2 ln x A n A ln x A n A n A n A + n B n B n A + n B n B n A n B n A n B n A n B n A (n A + n B ) (n A + n B ) µ A µ * A + RT ln x A + RT n A + n B n A + n + C B n B n A n B + n2 B n A n B (n A + n B ) 2 µ µ * A + RT ln x A + RT(0) + Cn 2 B (n A + n B ) 2 b. Write µ * A + RT ln x A + Cx2 B µ A µ * A + RT ln x A µ* A + RT ln γ A By comparison, RT ln γ A Cx 2 B or ln γ Cx2 B RT Thus γ e Cx2 B/RT when x B 0.
PHASES AND SOLUTIONS n This corresponds to a pure A. In a very dilute solution of A in B we also expect γ A. In that case µ * A lim x A 0 (µ A RT ln x A ) Substitution from above Therefore, µ * A lim x B (µ* A + Cx2 B ) µ* A + C µ µ * A + RT ln x A + C(x2 B ) µ* A + RT ln x A + RT ln γ A ln γ A C(x 2 B )/RT 0 when x B n Colligative Properties 5.48. From Eq. 5.5, 9 000 ln x 8.345 353.35 298.5 2.285 0 3 (2.830 0 3 3.354 0 3 ) 6.467 7.664.98 x 0.302 5.49. Rewriting Eq. 5.28 as P 2 (k ) c 2, the values for N 2 and O 2 may be substituted directly. We make the assumption that N 2 gives rise to 80% of the pressure and O 2 is responsible for the other 20%. Then, c(n 2 ) 0.80(0 325) Pa 2.7 0 8 mol dm 3 Pa.76 0 3 mol dm 3 c(o 2 ) 0.20(0 325) Pa.02 0 8 mol dm 3 Pa 2.07 0 4 mol dm 3 The total concentration is.967 0 3 mol dm 3. This value of the concentration approaches the value of the molality. We may then use the molal freezing point depression expression. The result is fus T (.86) (0.00 967) 0.003 66 K 5.50. Since the molecular weight is 60.06, c 0.00 mol dm 3 π crt n V RT 0.00 (8.345) (300.5) kpa 249.6 kpa
2 n CHAPTER 5 5.5. fus T K f m 2 2.7.50 3.255 K From Eq. 5.5, substituting a for x as discussed in the paragraph after Eq. 5.00, we have ln a fus H m R T fus Hθ T * f R T * f ln a 6009.5(3.255) 8.345(273.5) 2 0.035; a 0.969 x 55.6 55.6 +.5 0.9737 f a x 0.9690 0.9737 0.995 5.52. The molality of the solute is calculated from vap T K b m 2 m fus T 0.9 K K b 6.26 K m 0.44 m 0.44 mol kg The mass of solute per kilogram of solvent is 0.000 85 kg solute 0.50 kg bromobenzene 0.005 67 Then for the solute, 0.005 67 kg 0.44 mol and M 0.005 67 kg 0.44 mol 0.0394 kg mol 39.4 g mol 5.53. For the dissociation: A x B y xa z+ yb z Initial molalities: m 0 0 mol kg Molalities after dissociation: m αm xαm yαm mol kg Total molality m( α + xα + yα) mol kg Then i total molality initial molality α + xα + yα If ν x + y, i α + αν α ( ν) Then α i ν From our problem, fus T 273.50 273.4 0.036 K i fus T/K f m 0.036 K/[.86 (K m )0.00(m)].94
PHASES AND SOLUTIONS n 3 Since complete dissociation gives ν 2 for HCl, α i ν 0.94 0.94 The electrolyte therefore appears to be 94% dissociated. This extent of dissociation is only apparent because of the nonideality of the solution. 5.54. From Eq. 5.34, π n 2 RT V Thus π m 2 /M 2 V RT where m 2 is the mass of solute dissolved in volume V and M 2 is the molar mass. Since the equations are good only for dilute solutions, we try to find a limiting value of M 2. From the preceding, M 2 (RT/π) (m 2 /V) and so lim m 2 V 0 and lim m 2 πv 0. A plot of (m 2 /πv)/g dm 3 atm against (m 2 /V)/g dm 3 gives a limiting value of m 2 /πv. The values of (m 2 /πv)/g dm 3 atm corresponding to the listed values are 2.93 2.98 2.68 2.6.53 0.92 From the plot, the limiting value of m 2 /πv is about 2.9 g dm 3 atm. Thus M 2 RT m 2 V 0.082 atm dm 3 K mol 293.5 (K) 0 2.9 (g dm 3 atm ) 30 g mol The molar mass of sucrose is 342 g mol, so that there is an error of about 9%. Ignoring the lowest concentration point and extrapolating leads to an error of about 4%. However, since the slope is expected to be fairly close to zero at infinite dilution, a better result cannot be obtained without more low-concentration data. 5.55. From Eq. 5.22, M 2.856(K kg mol ) 3.78 g 0.646(K) 300.0 g 0.0362 kg mol 36.2 g mol 5.56. m 2 6.09 g/87.4 g mol 0.0325 mol m 250.0/8.05 3.88 mol 3.88 x 0.0325 + 3.88 From Eq. 5.25, 0.997 66
4 n CHAPTER 5 ln x ln 0.997 66 0.002 339 8.345 40 660 /T 2.6794 0 3 T 373.28 K 40 660 8.345 (/T /373.5) T 373.28 373.5 0.068 K From Eq. 5.26, /T /373.5 vap T K b m 2 0.54 K kg mol 0.0325 mol/0.2500 kg 0.0703 K The agreement is good. The difference between the two values indicates that the solution is sufficiently dilute for Eq. 5.26 to apply. 5.57. The Clapeyron equation (Eq. 5.9) may be used for this problem. Since the boiling point is given at a pressure for atm, it is appropriate to have the pressure of the gas given in atmospheres. Since mmhg Torr, log P/Torr log [P/atm atm (760 Torr) ] log P/atm + log atm (760 Torr) The derivative of this expression shows that it does not matter how pressure is expressed as long as we are considering only a change in pressure. Since d log P dt dp dt P 0.43429 log 0 e P dp dt 0.434 29 P dp dt, d dt [5.4672 427.3 T 369.3 T -2 ] We now can use the Clapeyron equation (Eq. 5.9). At the boiling point, P atm, T b 398.4 K. dp dt ( atm) 0.43429 [427.3/(398.4)2 + 6338.6/((398.4) 3 ] 2.094 0 2 atm K V m (liquid) 63.9 0 3 kg mol 0.89 kg dm 3 7.80 0 2 dm 3 mol V m (vapor) 63.9 0 3 kg mol 3.5 0 4 kg dm 3 202.86 dm3 mol V V m (vapor) V m (liquid) 202.86 0.0780 202.78 dm 3 mol Rearranging and substituting into Eq. 5.9, we have vap H m T V m dp dt 398.4 K(202.78 dm3 mol )(2.094 0 2 atm K ) 692 dm 3 atm mol 692 dm 3 atm mol 0 3 m 3 dm 3 0.325 kj atm m 3 7.4 kj mol
PHASES AND SOLUTIONS n 5 5.58. Addition of the moles per liter of the salts listed in Table 5.5 gives a.0989 molar solution. Assuming that there is sufficient positive charge to cancel the negative charges present, no calculation need be made to determine van t Hoff s i factor. Although the temperature of the ocean is variable, we can assume an average value of 25 C near the surface. Then π crt.0989 mol dm 3 (8.345 J K mol )(298 K) 2.72 0 3 J dm 3 Since J kg m 2 s 2 and Pa kg m s 2 π 2.72 0 3 kg m 2 s 2 dm 3 0 3 dm 3 /m 3 2.72 0 6 kg m s 2 2.72 MPa